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Question:I did an ANOVA test with an online calculator. I am really not sure how to interpret these results. The original story was that "psychologists are interested in studying the influence of media on aggressive behavior in kids. Among 15 subjects (n=15) five are exposed to violent video games, 5 to games and 5 to movies. After exposure, subjects were then observed by researchers and the # of aggressive behaviors of each subject was recorded by researchers yielding the following distribution of values. (I put those values into the test.)
I just dont know how to interpret these findings into words....... can anyone help me with this?
ANOVA: Results
The results of a ANOVA statistical test performed at 18:47 on 2-APR-2009
Source of Sum of d.f. Mean F
Variation Squares Squares
between 1485. 2 742.5 11.27
error 790.8 12 65.90
total 2276. 14
The probability of this result, assuming the null hypothesis, is 0.002
--------------------------------------------------------------------------------
Group A: Number of items= 5
25.0 38.0 39.0 42.0 47.0
Mean = 38.2
95% confidence interval for Mean: 30.29 thru 46.11
Standard Deviation = 8.17
Hi = 47.0 Low = 25.0
Median = 39.0
Average Absolute Deviation from Median = 5.20
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Group B: Number of items= 5
8.00 19.0 22.0 23.0 31.0
Mean = 20.6
95% confidence interval for Mean: 12.69 thru 28.51
Standard Deviation = 8.32
Hi = 31.0 Low = 8.00
Median = 22.0
Average Absolute Deviation from Median = 5.40
--------------------------------------------------------------------------------
Group C: Number of items= 5
5.00 11.0 14.0 18.0 26.0
Mean = 14.8
95% confidence interval for Mean: 6.890 thru 22.71
Standard Deviation = 7.85
Hi = 26.0 Low = 5.00
Median = 14.0
Average Absolute Deviation from Median = 5.60 Also... the question is also asked on the form: "What form of media seems to have the strongest effect on aggressive behavior and are the means significantly different?"
I have no idea how to actually interpret that and I am stuck, new to this.....

Answers:It is hard to tell you what media seems to have the effects, since you only refer to them as Group A, B and C, so you will have to look that up yourself.
First Question:
'What form of media seems to have the strongest effect on aggressive behavior?'
Just describe the data here.
For example, say that Group A has a mean of 38.2, with a highest value of 47 and a low of 25. Do the same for the other groups and say that it appears that Group A has the highest of all groups, and B is greater than C.
Second Question:
'Are the means significantly different?'
Basically, the ANOVA results are telling you that there is a significant difference in recorded aggressive behaviours for the three groups.
How can you tell that it is significant?
One way is to compare the alpha value you used in running the analysis. You have not reported it, but the most often used is 0.05, so I will assume it is that.
If you look at the p value you got in your results, you will see that it is equal to 0.002. Since 0.002 is smaller than 0.05, your result is significant.
As it turns out, your p value is lower than all three most often use alpha values (.10, .05, .01) so that's great :)
So what you can say is that there is significant differences in aggressive behaviour after exposure to violent video games, games and movies, with F(2,12)=11.27, p<0.05 with Group A showing greater aggressive behaviour than both B and C, and B greater than C.
You could also mention the confidence intervals.
These show that you can be 95% confident that the true population value falls between these values and because the range of these values do not include 0, the result is statistically significant.

Question:I pretty much have zero knowledge in using SPSS, and very little knowledge about statistics, etc. I have to do this data analysis using regression (linear). So the result that has been generated is this:
the value of R= .620a
R square= .385
adjusted R square=.350
the coefficients:
(Constant)=.862 (t)
Pay= -.672 (t)
Personal factors= 1.445 (t)
Job design and characteristics=1.956 (t)
Workplace spirituality= 3.042 (t)
Physical structure of workplace= 1.362 (t)
note: the dependent variable is employee performance.
I think it is not a good analysis isn't it. Can you please help me by giving comments? what does the R or R sq value mean in this case? how can interpret the results with regards to the variables?
what should i do when the R square is too low?

Answers:There is a better category for you to post your question: "Math" or "Homeworrk."
Social Science category is about people, we leave the mathmatical equations to the experts in math category.
They love that stuff. Good luck.

Question:I have a lab I am doing for a chemistry class and I dont quite understand this question
Determine the concentration of your diluted solution. Assume the concentration of your drink was 0.15M and you make a tenfold dilution. (Write your data, substitute the values in the formula for density and show your stepwise calculation).
I know you are supposed to use Mi*Vi=Mf*Vf but im not sure how to start with the question we had started with 1mL then added water untill it hit the 10 mL mark

Question:I did a 10 question pre & post survey of the same 10 people. The mean difference is -1.3, Degrees of freedom= 9, standard error of the mean of d= .86, t-statistic for paired data = -1.51, Critical value for alpha .05 would be 2.262.
Now I am not sure how to know whether I accept or reject the null hypothesis. Not even sure what the null hypothesis should be.
My research is whether or not medication improves the school performance of ADHD children. Pre-Survey (before meds) and after survey (while on meds).
I have all these numbers and don't know how to make since of them. Guess I should've paid more attention in that statistics class!
Please help.

Answers:It's very difficult to answer what the null hypothesis should be. If you don't know what you're comparing, then why do the test? I assume you are comparing pre and post test results. Your null hypothesis would most likely be that the test results are the same for both tests. It also sounds like you're preforming a one-sided test on what could be a two-sided test. You would reject any statistic that is above 2.262 or below -2.262 and this would give you an alpha of .05 (.025 on each tail). Basically it looks like you found a two-sided significant value (2.262) but think you're doing a one-sided test. If you're looking to see if there's a significant difference period, then you'd use the two-sided test. If you want to see if the second test scored better, then use the one - sided. (It'd be a lot easier to explain this if I knew exactly how you did the math). If you go above the significant value or below it then you reject the null. Regardless, this won't change your result in this case since your value is -1.51. Basically, any statistic outside of alpha means you reject the null because you have a statistically significant reason to say that the data in the first test is different than the data in the second test. You are making sure there is a significant difference in the data in order to say the medication made a change. Based on your results, you aren't seeing a change since the results. You would not reject the null hypothesis. This means that, based on your results, this drug does not alter test results in patients with ADHD. And yes, you should have paid more attention in statistics class. This stuff really isn't that difficult but you seem so lost. If you plan on doing more of this type of thing I'd suggest taking a stats class again or at least getting a stats textbook and look at it. Also, make sure this is the correct test and that you did it correctly. Based on what you've said, I'm assuming that your resulting t-statistic is correct and that you're using the correct test. It sounds like you have (comparing two results given what it's for) but make sure.

From Youtube

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